Solution Manual for Abstract Algebra An Introduction 3rd Edition by Hungerfor Chapter © Cengage Learning All rights reserved No distribution allowed without express authorization Arithmetic in Z Revisited 1.1 The Division Algorithm (a) q = 4, r = (b) q = 0, r = (c) q = −5, r = (a) q = −9, r = (b) q = 15, r = 17 (c) q = 117, r = 11 (a) q = 6, r = 19 (b) q = −9, r = 54 (c) q = 62720, r = 92 (a) q = 15021, r = 132 (b) q = −14940, r = 335 (c) q = 39763, r = 3997 Suppose a = bq + r, with ≤ r < b Multiplying this equation through by c gives ac = (bc)q + rc Further, since ≤ r < b, it follows that ≤ rc < bc Thus this equation expresses ac as a multiple of bc plus a remainder between and bc − Since by Theorem 1.1 this representation is unique, it must be that q is the quotient and rc the remainder on dividing ac by bc When q is divided by c, the quotient is k, so that q = ck Thus a = bq + r = b(ck) + r = (bc)k + r Further, since ≤ r < b, it follows (since c ≥ 1) than ≤ r < bc Thus a = (bc)k + r is the unique representation with ≤ r < bc, so that the quotient is indeed k Answered in the text Any integer n can be divided by with remainder r equal to 0, 1, or Then either n = 4k, 4k + 1, 4k + or 4k + 3, where k is the quotient If n = 4k or 4k + then n is even Therefore if n is odd then n = 4k + or 4k + We know that every integer a is of the form 3q, 3q + or 3q + for some q In the last case a3 = (3q + 2)3 = 27q3 + 54q2 + 36q + = 9k + where k = 3q3 + 6q2 + 4q Other cases are similar 10 Suppose a = nq + r where ≤ r < n and c = nq' + r' where < r' < n If r = r' then a – c = n(q – q') and k = q – q' is an integer Conversely, given a – c = nk we can substitute to find: (r – r') = n(k – q + q') Suppose r ≥ r' (the other case is similar) The given inequalities imply that ≤ (r – r') < n and it follows that ≤ (k – q + q') < 101 and we conclude that k – q + q' = Therefore r – r' = 0, so that r = r' as claimed Not For Sale © 2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use Solution Manual for Abstract Algebra An Introduction 3rd Edition by Hungerfor Not For Sale Arithmetic in Z Revisited 11 Given integers a and c with c ≠ Apply Theorem 1.1 with b = |c| to get a = |c| q + r where 0 ≤ r < |c| Let q = q0 if c > and q = –q0 if c < Then a = cq + r as claimed The uniqueness is proved as in Theorem 1.1 Divisibility (a) (b) (c) (d) 11 (e) (f) 17 (g) 592 (h) If b | a then a = bx for some integer x Then a = (–b)(–x) so that (–b) | a The converse follows similarly Answered in the text (a) Given b = ax and c = ay for some integers x, y, we find b + c = ax + ay = a(x + y) Since x + y is an integer, conclude that a | (b + c) (b) Given x and y as above we find br + ct = (ax)r + (ay)t = a(xr + yt) using the associative and distributive laws Since xr + yt is an integer we conclude that a | (br + ct) Since a | b, we have b = ak for some integer k, and a = Since b | a, we have a = bl for some integer l, and b = Thus a = bl = (ak)l = a(kl) Since a = 0, divide through by a to get = kl But this means that k = ±1 and l = ±1, so that a = ± b Given b = ax and d = cy for some integers x, y, we have bd = (ax)(cy) = (ac)(xy) Then ac | bd because xy is an integer Clearly (a, 0) is at most |a| since no integer larger than |a| divides a But also |a| | a, and |a| | since any nonzero integer divides Hence |a| is the gcd of a and If d = (n, n + 1) then d | n and d | (n + 1) Since (n + 1) – n = we conclude that d | (Apply Exercise 4(b).) This implies d = 1, since d > No, ab need not divide c For one example, note that | 12 and | 12, but · = 24 does not divide 12 10 Since a | a and a | we have a | (a, 0) If (a, 0) = then a | forcing a = ±1 (b) 1, 2, or Generally if d = (n, n + c) then d | n and d | (n + c) 11 (a) or Since c is a linear combination of n and n+c, conclude that d | c 12 (a) False (ab, a) is always at least a since a | ab and a | a (b) False For example, (2, 3) = and (2, 9) = 1, but (3, 9) = (c) False For example, let a = 2, b = 3, and c = Then (2, 3) = = (2, 9), but (2 · 3, 9) = © 2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use © Cengage Learning All rights reserved No distribution allowed without express authorization 1.2 Solution Manual for Abstract Algebra An Introduction 3rd Edition by Hungerfor 1.2 Divisibility 13 (a) Suppose c | a and c | b Write a = ck and b = cl Then a = bq + r can be rewritten ck = (cl)q + r, so that r = ck − clq = c(k − lq) Thus c | r as well, so that c is a common divisor of b and r (b) Suppose c | b and c | r Write b = ck and r = cl, and substitute into a = bq + r to get a = ckq + cl = c(kq + l) Thus c | a, so that c is a common divisor of a and b (c) Since (a, b) is a common divisor of a and b, it is also a common divisor of b and r, by part (a) If (a, b) is not the greatest common divisor (b, r) of b and r, then (a, b) > (b, r) Now, consider (b, r) By part (b), this is also a common divisor of (a, b), but it is less than (a, b) This is a contradiction Thus (a, b) = (b, r) © Cengage Learning All rights reserved No distribution allowed without express authorization 14 By Theorem 1.3, the smallest positive integer in the set S of all linear combinations of a and b is exactly (a, b) (a) (6, 15) = (b) (12, 17)=1 15 (a) This is a calculation (b) At the first step, for example, by Exercise 13 we have (a, b) = (524, 148) = (148, 80) = (b, r) The same applies at each of the remaining steps So at the final step, we have (8, 4) = (4, 0); putting this string of equalities together gives (524, 148) = (148, 80) = (80, 68) = (68, 12) = (12, 8) = (8, 4) = (4, 0) But by Example 4, (4, 0) = 4, so that (524, 148) = (c) 1003 = 56 · 17 + 51, 56 = 51 · + 5, 51 = · 10 + 1, = · + Thus (1003, 56) = (1, 0) = (d) 322 = 148 · + 26, 148 = 26 · + 18, 26 = 18 · + 8, 18 = · + 2, = · + 0, so that (322, 148) = (2, 0) = (e) 5858 = 1436 · + 114, 1436 = 114 · 12 + 68, 114 = 68 · + 46, 68 = 46 · + 22, 46 = 22 · + 2, 22 = · 11 + 0, so that (5858, 1436) = (2, 0) = (f) 68 = 148 − (524 − 148 · 3) = −524 + 148 · (g) 12 = 80 − 68 · = (524 − 148 · 3) − (−524 + 148 · 4) · = 524 · − 148 · (h) = 68 − 12 · = (−524 + 148 · 4) − (524 · − 148 · 7) · = −524 · 11 + 148 · 39 (i) = 12 − = (524 · − 148 · 7) − (−524 · 11 + 148 · 39) = 524 · 13 − 148 · 46 (j) Working the computation backwards gives = 1003 · 11 − 56 · 197 16 Let a = da1 and b = db1 Then a1 and b1 are integers and we are to prove: (a1, b1) = l By Theorem 1.3 there exist integers u, v such that au + bv = d Substituting and cancelling we find that a1u + b1v = l Therefore any common divisor of a1 and b1 must also divide this linear combination, so it divides Hence (a1, b1) = 17 Since b | c, we know that c = bt for some integer t Thus a | c means that a | bt But then Theorem 1.4 tells us, since (a, b) = 1, that a | t Multiplying both sides by b gives ab | bt = c 18 Let d = (a, b) so there exist integers x, y with ax + by = d Note that cd | (ca, cb) since cd divides ca and cb Also cd = cax + cby so that (ca, cb) | cd Since these quantities are positive we get cd = (ca, cd) 19 Let d = (a, b) Since b + c = aw for some integer w, we know c is a linear combination of a and b so that d | c But then d | (b, c ) = forcing d = Similarly (a, c ) = Not For Sale © 2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use Solution Manual for Abstract Algebra An Introduction 3rd Edition by Hungerfor Not For Sale Arithmetic in Z Revisited 20 Let d = (a, b) and e = (a, b + at) Since b + at is a linear combination of a and b, d | (b + at) so that d | e Similarly since b = a(–t) + (b + at) is a linear combination of a and b + at we know e | b so that e | d Therefore d = e 21 Answered in the text 23 Define the powers bn recursively as follows: b1 = b and for every n ≥ 1, bn + = b bn By hypothesis (a, b1) = Given k ≥ 1, assume that (a, bk) = Then (a, bk + 1) = (a, b bk) = by Exercise 24 This proves that (a, bn) = for every n ≥ 24 Let d = (a, b) If ax + by = c for some integers x, y then c is a linear combination of a and b so that d | c Conversely suppose c is given with d | c, say c = dw for an integer w By Theorem 1.3 there exist integers u, v with d = au + bv Then c = dw = auw + bvw and we use x = uw and y = vw to solve the equation 25 (a) Given au + bv = suppose d = (a, b) Then d | a and d | b so that d divides the linear combination au + bv = Therefore d = (b) There are many examples For instance if a = b = d = u = v = then (a, b) = (1, 1)= while d = au + bv = + = 26 Let d Since Then Then = (a, b) and express a = da1 and b = db1 for integers a1, b1 By Exercise 16, (a1, b1) = a | c we have c = au = da1u for some integer u Similarly c = bv = db1v for some integer v a1u = c/d = b1V and Theorem 1.5 implies that a1 | v so that v = a1w for some integer w c = da1b1w so that cd = d2a1b1w = abw and ab | cd 27 Answered in the text 28 Suppose the integer consists of the digits an an−1 a1 a0 Then the number is equal to n n ak 10k = k=0 n ak (10k − 1) + k=0 ak k=0 Now, the first term consists of terms with factors of the form 10k − 1, all of which are of the form n 999 99, which are divisible by 3, so that the first term is always divisible by Thus k=0 ak 10k n is divisible by if and only if the second term k=0 ak is divisible by But this is the sum of the digits 29 This is almost identical to Exercise 28 Suppose the integer consists of the digits an an−1 a1 a0 Then the number is equal to n n ak 10k = k=0 n ak (10k − 1) + k=0 ak k=0 Now, the first term consists of terms with factors of the form 10k − 1, all of which are of the form n 999 99, which are divisible by 9, so that the first term is always divisible by Thus k=0 ak 10k n is divisible by if and only if the second term k=0 ak is divisible by But this is the sum of the digits © 2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use © Cengage Learning All rights reserved No distribution allowed without express authorization 22 Let d = (a, b, c) Claim: (a, d) = l [Proof: (a, d) divides d so it also divides c Then (a, d) | (a, c) = so that (a, d)= 1.] Similarly (b, d)= But d | ab and (a, d) = so that Theorem 1.5 implies that d | b Therefore d = (b d) = Solution Manual for Abstract Algebra An Introduction 3rd Edition by Hungerfor 1.2 Divisibility 30 Let S = {a1x1 + a2x2 + … + anxn : x1 x2, , x are integers} As in the proof of Theorem 1.3, S does contain some positive elements (for if aj ≠ then aj2 ∈ S is positive) By the Well Ordering Axiom this set S contains a smallest positive element, which we call t Suppose t = a1u1 + a2u2 + … + anun for some integers uj Claim t = d The first step is to show that t | a1 By the division algorithm there exist integers q and r such that a1 = tq + r with ≤ r < t Then r = a1 – tq = a1(1 – u1q) + a2(–u2q) + … + an(–unq) is an element of S Since r < t (the smallest positive element of S), we know r is not positive Since r ≥ the only possibility is r = Therefore a1 = tq and t | a1 Similarly we have t | aj for each j, and t is a common divisor of a1, a2,…, an Then t ≤ d by definition On the other hand d divides each aj so d divides every integer linear combination of a1, a2, , an © Cengage Learning All rights reserved No distribution allowed without express authorization In particular, d | t Since t > this implies that d ≤ t and therefore d = t 31 (a) [6, 10] = 30; [4, 5, 6, 10] = 60; [20, 42] = 420, and [2, 3, 14, 36, 42] = 252 (b) Suppose | t for i = 1, 2, , k, and let m = [a1 , a2 , , ak ] Then we can write t = mq + r with ≤ r < m For each i, | t by assumption, and | m since m is a common multiple of the Thus | (t − mq) = r Since | r for each i, we see that r is a common multiple of the But m is the smallest positive integer that is a common multiple of the ; since ≤ r < m, the only possibility is that r = so that t = mq Thus any common multiple of the is a multiple of the least common multiple 32 First suppose that t = [a, b] Then by definition of the least common multiple, t is a multiple of both a and b, so that t | a and t | b If a | c and b | c, then c is also a common multiple of a and b, so by Exercise 31, it is a multiple of t so that t | c Conversely, suppose that t satisfies the conditions (i) and (ii) Then since a | t and b | t, we see that t is a common multiple of a and b Choose any other common multiple c, so that a | c and b | c Then by condition (ii), we have t | c, so that t ≤ c It follows that t is the least common multiple of a and b da1 db1 = da1 b1 Since a and b are 33 Let d = (a, b), and write a = da1 and b = db1 Write m = ab d = d both positive, so is m, and since m = da1 b1 = (da1 )b1 = ab1 and m = da1 b1 = (db1 )a1 = ba1 , we see that m is a common multiple of a and b Suppose now that k is a positive integer with a | k and b | k Then k = au = bv, so that k = da1 u = db1 v Thus kd = a1 u = b1 v By Exercise 16, (a1 , b1 ) = 1, so that a1 | v, say v = a1 w Then k = db1 v = db1 a1 w = mw, so that m | k Thus ab m ≤ k It follows that m is the least common multiple But by construction, m = (a,b) = ab d 34 (a) Let d = (a, b) Since d | a and d | b, it follows that d | (a + b) and d | (a − b), so that d is a common divisor of a + b and a − b Hence it is a divisor of the greatest common divisor, so that d = (a, b) | (a + b, a − b) (b) We already know that (a, b) | (a+b, a−b) Now suppose that d = (a+b, a−b) Then a+b = dt and a − b = du, so that 2a = d(t + u) Since a is even and b is odd, d must be odd Since d | 2a, it follows that d | a Similarly, 2b = d(t − u), so by the same argument, d | b Thus d is a common divisor of a and b, so that d | (a, b) Thus (a, b) = (a + b, a − b) (c) Suppose that d = (a + b, a − b) Then a + b = dt and a − b = du, so that 2a = d(t + u) Since a and b are both odd, a + b and a − b are both even, so that d is even Thus a = d2 (t + u), so that d2 | a Similarly, d2 | b, so that d2 = (a+b,a−b) | (a, b) | (a + b, a − b) Thus (a, b) = (a+b,a−b) 2 or (a, b) = (a + b, a − b) But since (a, b) is odd and (a + b, a − b) is even, we must have (a+b,a−b) = (a, b), or 2(a, b) = (a + b, a − b) Not For Sale © 2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use Solution Manual for Abstract Algebra An Introduction 3rd Edition by Hungerfor Not For Sale Arithmetic in Z Revisited Primes and Unique Factorization (a) 24 · 32 · · (c) · · 4567 (d) 23 · · · · 11 · 13 · 17 (b) −5 · · 67 √ (a) Since 25 − = 31, and 31 < 6, we need only check divisibility by the primes 2, 3, and Since none of those divides 31, it is prime √ (b) Since 27 − = 127, and 127 < 12, we need only check divisibility by the primes 2, 3, 5, 7, and 11 Since none of those divides 127, it is prime (c) 211 − = 2047 = 23 · 89 They are all prime The pairs are {3, 5}, {5, 7}, {11, 13}, {17, 19}, {29, 31}, {41, 43}, {59, 61}, {71, 73}, {101, 103}, {107, 109}, {137, 139}, {149, 151}, {179, 181}, {191, 193}, {197, 199} (a) Answered in the text These divisors can be listed as 2j.3k for ≤ j ≤ s and ≤ k ≤ t (b) The number of divisors equals (r + l)(s + l)(t + 1) The possible remainders on dividing a number by 10 are 0, 1, 2, , If the remainder on dividing p by 10 is 0, 2, 4, 6, or 8, then p is even; since p > 2, p is divisible by in addition to and itself and cannot be prime If the remainder is 5, then since p > 5, p is divisible by in addition to and itself and cannot be prime That leaves as possible remainders only 1, 3, 7, and Since p | (a + bc) and p | a, we have a = pk and a + bc = pl, so that pk + bc = pl and thus bc = p(l − k) Thus p | bc By Theorem 1.5, either p | b or p | c (or both) (a) As polynomials, xn − = (x − 1)(xn−1 + xn−2 + · · · + x + 1) (b) Since 22n · 3n − = (22 · 3)n − = 12n − 1, by part (a), 12n − is divisible by 12 − = 11 If p is a prime and p = rs then by the definition r, s must lie in {1, –1, p, –p} Then either r = ±1 or r = ±p and s = p/r = ±1, Conversely if p is not a prime then it has a divisor r not in {1, –1, p, –p} Then p = rs for some integer s If s equals ±1 or ±p then r = p/s would equal ±p or +1, contrary to assumption This r, s provides an example where the given statement fails 10 Assume first that p > If p is a prime then (a, p) is a positive divisor of p, so that (a, p) = or p If (a, p) = p then p | a Conversely if p is not a prime it has a divisor d other than ±1 and ±p We may change signs to assume d > Then (p, d) = d ≠ l Also p | d since otherwise p | d and d = p implies d = p Then a = d provides an example where the required statement fails Finally if p < apply the argument above to –p © 2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use © Cengage Learning All rights reserved No distribution allowed without express authorization 1.3 Solution Manual for Abstract Algebra An Introduction 3rd Edition by Hungerfor 1.3 Primes and Unique Factorization 11 Since p | a − b and p | c − d, also p | (a − b) + (c − d) = (a + c) − (b + d) Thus p is a divisor of (a + c) − (b + d); the fact that p is prime means that it is a prime divisor © Cengage Learning All rights reserved No distribution allowed without express authorization 12 Since n > Theorem 1.10 implies that n equals a product of primes We can pull out minus signs to see that n = p1 p2 … pr where each pj is a positive prime Re-ordering these primes if necessary, to assume p ≤ p ≤ … ≤ p r For the uniqueness, suppose there is another factorization n = q q 2…q s for some positive primes qj with q ≤ q … ≤ q s By theorem 1.11 we know that r = s and the p j’s are just a re-arrangement of the q js Then p is the smallest of the p j’s, so it also equals the smallest of the q j’s and therefore p = q We can argue similarly that p = q 2, …, p r = q r (This last step should really be done by a formal proof invoking the Well Ordering Axiom.) 13 By Theorem 1.8, the Fundamental Theorem of Arithmetic, every integer except and ±1 can be written as a product of primes, and the representation is unique up to order and the signs of the primes Since in our case n > is positive and we wish to use positive primes, the representation is unique up to order So write n = q1 q2 qs where each qi > is prime Let p1 , p2 , , pr be the distinct primes in the list Collect together all the occurrences of each pi , giving ri copies of pi , i.e pri i 14 Suppose d | p so that p = dt for some integer t The hypothesis then implies that p | d or p | t If p | d then (applying Exercise 1.2.5) d = ±p Similarly if p | t then, since we know that t | p, we get t = +p, and therefore d = ±1 15 Apply Corollary 1.9 in the case a1 =a2 = … = an to see that if p | an then p | a Then a = pu for some integer u, so that an = pnun and pn | an 16 Generally, p | a and p | b if and only if p | (a, b), as in Corollary 1.4 Then the Exercise is equivalent to: (a, b) = if and only if there is no prime p such that p | (a, b) This follows using Theorem 1.10 17 First suppose u, v are integers with (u, v) = Claim (u2, v2) = For suppose p is a prime such that p | u2 and p | v2 Then p | u and p | v (using Theorem 1.8), contrary to the hypothesis (u, v) = Then no such prime exists and the Claim follows by Exercise Given (a, b) = p write a = pa1 and b = pb1 Then (a1, b1) = by Exercise 1.2.16 Then (a2, b2) = (p2a12, p2b12) = p2(a12, b12), using Exercise 1.2.18 By the Claim we conclude that (a2, b2) = p2 18 The choices p = 2, a = b = 0, c = d = provide a counterexample to (a) and (b) (c) Since p | (a2 + b2) – a.a = b2, conclude that p | b by Theorem 1.8 19 If ri ≤ si for every i, then b = ps11 ps22 pskk = pr11 ps11 −r1 pr22 ps22 −r2 prkk pskk −rk = (pr11 pr22 prkk ) · ps11 −r1 ps22 −r2 psk2 −rk = a · ps11 −r1 ps22 −r2 psk2 −rk Since each si − ri ≥ 0, the second factor above is an integer, so that a | b Now suppose a | b, and consider pri i Since this is composed of factors only of pi , it must divide psi i , since pi pj for i = j Thus pri i | psi i Clearly this holds if ri ≤ si , and also clearly it does not hold if ri > si , since then pri i > psi i Not For Sale © 2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use Solution Manual for Abstract Algebra An Introduction 3rd Edition by Hungerfor Not For Sale Arithmetic in Z Revisited 20 (a) The positive divisors of a are the numbers d = p1m1p2m2…pkmk where the exponents mj satisfy ≤ mj ≤ rj for each j = 1, 2,, , k This follows from unique factorization If d also divides b we have ≤ mj ≤ sj for each i = 1, 2, k Since nj = min{rj, sj} we see that the positive common divisors of a and b are exactly those numbers d = p1m1p2m2 … pkmk where ≤ mj ≤ nj for each j = 1, 2, , k Then (a, b) is the largest among these common divisors, so it equals p1n1p2n2…pknk (b) For [a, b] a similar argument can be given, or we can apply Exercise 1.2.31, noting that max{r, s} = r + s – min{r, s} for any positive numbers r, s 22 If every ri is even it is easy to see that n is a perfect square Conversely suppose n is a square First consider the special case n = pr is a power of a prime If pr = m2 is a square, consider the prime factorization of m By the uniqueness (Theorem 1.11), p is the only prime that can occur, so m = ps for some s, and pr = m2 = p2s Then r = 2s' is even Now for the general case, suppose n = m2 is a perfect square If some ri is odd, express n = piri k where k is the product of the other primes involved in n Then piri and k are relatively prime and Exercise 13 implies that piri is a perfect square By the special case, ri is even 23 Suppose a = pr11 pr22 prkk and b = ps11 ps22 pskk where the pi are distinct positive primes and ri ≥ 0, 2rk 2sk 2r2 2s2 si ≥ Then a2 = p2r and b2 = p2s p2 pk p2 pk Then using Exercise 19 (twice), we have a | b if and only if ri ≤ si for each i if and only if 2ri ≤ 2si for each i if and only if a2 | b2 24 This is almost identical to the previous exercise If n > is an integer, suppose a = pr11 pr22 prkk and b = ps11 ps22 pskk where the pi are distinct positive primes and ri ≥ 0, si ≥ Then an = nr2 k ns2 k pnr pnr and b2 = pns pns p2 p2 k k Then using Exercise 19 (twice), we have a | b if and only if ri ≤ si for each i if and only if nri ≤ nsi for each i if and only if an | bn 25 The binomial coefficient p k is p k = p! p · (p − 1) · · · (p − k + 1) = k!(p − k)! k(k − 1) · · · Now, the numerator is clearly divisible by p The denominator, however, consists of a product of integers all of which are less than p Since p is prime, none of those integers (except 1) divide p, so the product cannot have a factor of p (to make this more precise, you may wish to write the denominator as a product of primes and note that p cannot appear in the list) 26 Claim: Each Ak = (n + 1)! + k is composite, for k = 2, 3, , n + Proof Since k ≤ n + we have k | (n + 1)! and therefore k | Ak Then Ak is composite since I < k < Ak 27 By the division algorithm p = 6k + r where ≤ r < Since p > is prime it is not divisible by or 3, and we must have r = or If p = 6k + then p2 = 36k2 + 12k + and p2 + = 36k2 + 12k + is a multiple of Similarly if p = 6k + then p2 +2 = 36k2 + 60k + 27 is a multiple of So in each case, p2 + is composite © 2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use © Cengage Learning All rights reserved No distribution allowed without express authorization 21 Answered in the text Solution Manual for Abstract Algebra An Introduction 3rd Edition by Hungerfor 1.3 Primes and Unique Factorization 28 The sums in question are: + + + … + 2n When n = the sum is 255 = 3.5.17 and when n = the sum is 511 = 7.73 Therefore the assertion is false The interested reader can verify that this sum equals 2n+1 – These numbers are related to the “Mersenne primes” 29 This assertion follows immediately from the Fundamental Theorem 1.11 © Cengage Learning All rights reserved No distribution allowed without express authorization 30 (a) If a2 = 2b2 for positive integers a, b, compare the prime factorizations on both sides The power of occurring in the factorization of a2 must be even (since it is a square) The power of occurring in 2b2 must be odd By the uniqueness of factorizations (The Fundamental Theorem) these powers of must be equal, a contradiction (b) If is rational it can be expressed as a fraction ab for some positive integers a, b Clearing denominators and squaring leads to: a2 = 2b2, and part (a) applies 31 The argument in Exercise 20 applies More generally see Exercise 27 below 32 Suppose all the primes can be put in a finite list p1, p2,…, pk and consider N = p1 p2 …pk + None of these pj can divide N (since can be expressed as a linear combination of pj and N) But N > so N must have some prime factor p (Theorem 1.10) This p is a prime number not equal to any of the primes in our list, contrary to hypothesis 33 Suppose n is composite, and write n = rs where < r, s < n Then, as you can see by multiplying it out, 2n − = (2r − 1) 2s(r−1) + 2s(r−2) + 2s(r−3) + · · · + 2s + Since r > 1, it follows that 2r > Since s > 1, we see that 2s + > 1, so that the second factor must also be greater than So 2n − has been written as the product of two integers greater than one, so it cannot be prime 34 Proof: Since n > we know that n! – > so it has some prime factor p If p ≤ n then p | n!, contrary to the fact that p | n! Therefore n < p < n! 35 We sketch the proof (b) Suppose a > (What if a < 0?), rn = a and r = u/v where u, v are integers and v > Then un = avu If p is a prime let k be the exponent of p occurring in a (that k +1 is: pk | a and p | a ) The exponents of p occurring in un and in must be multiples of n, so unique factorization implies k is a multiple of n Putting all the primes together we conclude that a = bn for some integer b 36 If p is a prime > then | p and | p, so by Exercise 1.2.34 we know 24 | p2 – Similarly 24 | (q2 – 1) so that p2 – q2 = (p2 – 1) – (q2 – 1) is a multiple of 24 Not For Sale © 2013 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use © Cengage Learning All rights reserved No distribution allowed without express authorization Solution Manual for Abstract Algebra An Introduction 3rd Edition by Hungerfor Not For Sale